- ✓$\Delta G = 0$
- B$\Delta S = 0$
- C$\Delta H = 0$
- D$\Delta U = 0$
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$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$, the $\mathrm{K}_{\mathrm{p}}$ for the process is $\mathrm{x} \times 10^{-1} \cdot \mathrm{x}=$ $\qquad$
[Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
$CN^+, CN^-, NO$ and $CN$
Which one of these will have the highest bond order?
$[A]$ $HClO_4$ is more acidic than $\mathrm{HClO}$ because of the resonance stabilization of its anion
$[B]$ $HClO_4$ is formed in the reaction between $Cl_2$ and $H_2 O$
$[C]$ The central atom in both $HClO_4$ and $HClO$ is $s p^3$ hybridized
$[D]$ The conjugate base of $HClO_4$ is weaker base than $H_2 O$